Download presentation

Presentation is loading. Please wait.

1
Solving Gas Law Problems Copyright Sautter 2003

2
**SOLVING GAS LAW PROBLEMS**

BOYLE’S LAW CHARLES LAW GAY-LUSSAC’S LAW THE COMBINED GAS LAW THE IDEAL GAS LAW DALTON’S LAW GRAHAM’S LAW OF DIFFUSION

3
**GAS LAW FORMULAE BOYLE’S LAW: P1 x V1 = P2 x V2 CHARLES LAW:**

V1 / T1 = V2 / T2 GAY-LUSSAC’S LAW: P1 / T1 = P2 / T2 KELVIN = DEGREES CELSIUS COMBINED GAS LAW (P1 x V1 ) / T1 = (P2 x V2 ) / T2 DALTON’S LAW P TOTAL = PGAS A P GAS B P GAS C + P …….

4
**GAS LAW FORMULAE (CONT’D)**

DALTON’S LAW (CONT’D) PGAS A = (N GAS A / N TOTAL) x PTOTAL AVOGADRO’S HYPOTHESIS “EQUAL VOLUMES OF DIFFERENT GASES, AT THE SAME TEMPERATURE AND PRESSURE, CONTAIN EQUAL MOLES” UNIVERSAL GAS LAW P x V = N x R x T

5
**GAS LAW FORMULAE (CONT’D)**

GRAHAM’S LAW OF DIFFUSION v2 / v1 = ( m1 / m2)1/2 v = average molecular velocity m = molecular mass ONE MOLE OF ANY GAS OCCUPIES 22.4 LITERS AT STP CONDITIONS Liters divide by moles Liters multiply by moles

6
**SOLVING BOYLE’S LAW PROBLEMS**

WHAT IS THE VOLUME OF 500 ML OF NEON GAS AT 2.0 ATMS OF PRESSURE WHEN ITS PRESSURE IS CHANGED TO 2090 MM OF HG ? SOLUTION: P1 x V1 = P2 x V2 , P1 = 2.0 ATM V1 = 500 ML, P2 = MM / 760 = 2.75 ATM V2 = ( P1 x V1) / P2 V2 = (2.0 x 500) / = 364 ML NOTE: BOYLE’S LAW IS INVERSE, AS PRESSURE INCREASES, VOLUME DECREASES.

7
**SOLVING BOYLE’S LAW PROBLEMS**

IF 6.0 LITERS OF OXYGEN AT 1140 MM OF HG IS REDUCED TO A VOLUME OF 2000 ML, WHAT IS THE NEW PRESSURE OF THE GAS ? SOLUTION: P1 x V1 = P2 x V2 , P1 = 1140 MM V1 = 6.0 L, V2 = ML / 1000 = 2.0 L P2 = ( P1 x V1) / V2 P2 = (1140 x 6.0) / 2.0 = MM OF HG

8
**SOLVING CHARLES LAW PROBLEMS**

WHAT IS THE VOLUME OF HYDROGEN WHEN 300 ML ARE HEATED FROM 35 CELSIUS TO 80 CELSIUS ? SOLUTION: V1 / T1 = V2 / T2 , V1 = 300 ML KELVIN = DEGREES CELSIUS T1 = ( ) = 308 K, T2 = ( ) = 353 K V2 = (V1 x T2 ) / T1 V2 = (300 x 353) / 308 = 344 ml

9
**SOLVING CHARLES LAW PROBLEMS**

A 500 ml sample of carbon dioxide is reduced to 350 ml by cooling. If the original temperature has 300 K, what is the new temperature in degrees Celsius ? SOLUTION: V1 / T1 = V2 / T2 , V1 = 500 ML, V2 = 350 ML T1 = 300 K T2 = (V2 x T1 ) / V1 T2 = (350 x 300) / 500 = 210 K KELVIN = DEGREES CELSIUS 210 = C0 , C0 = - 63

10
**SOLVING GAY-LUSSAC LAW PROBLEMS**

WHAT IS THE PRESSURE OF A CONFINED GAS WITH AN ORIGINAL PRESSURE OF 3.0 ATM AND A TEMPERATURE OF 200K IF THE TEMPERATURE IS INCREASED TO 1000 C0 ? SOLUTION: P1 / T1 = P2 / T2 , P1 = 3.0 ATM, T1 = 200 K T2 = C0 KELVIN = DEGREES CELSIUS K = = K P2 = (P1 x T2 ) / T1 P2 = (3.0 x 1273) / 200 = ATM

11
**SOLVING GAY-LUSSAC LAW PROBLEMS**

A SAMPLE OF CHLORINE IS RAISED TO 1140 MM OF HG FROM A PRESSURE OF 0.50 ATM. IF THE ORIGINAL TEMPERATURE WAS 500 K, WHAT IS THE NEW TEMPERATURE IN CELSIUS ? SOLUTION: P1 / T1 = P2 / T2 , P1 = 0.50 ATM, P2 = 1140 / 760 = 1.5 ATM T1 = 500 K T2 = (P2 x T1 ) / P1 T2 = (1.5 x 500) / 0.50 = K KELVIN = DEGREES CELSIUS 1500 = C0 , C0 = 1227

12
**SOLVING COMBINED LAW PROBLEMS**

WHAT IS THE NEW VOLUME OF 650 ML OF NITROGEN AT 273 K AND 2.0 ATMS WHEN IT IS HEATED TO 819 K AND REDUCED TO 1.0 ATM PRESSURE ? SOLUTION: (P1 x V1 ) / T1 = (P2 x V2 ) / T2, P1 = 2.0 ATM, P2 = 1.0 ATM T1 = 273 K, T2 = 819 K V1 = 650 ml V2 = (P1 x V1 x T2 ) / (P2 x T1 ) V2 = (2.0 x x 819) / (1.0 x 273) = ml

13
**SOLVING COMBINED LAW PROBLEMS**

WHAT IS THE NEW PRESSURE OF 850 ML OF ARGON AT 1092 K AND 5.0 ATMS WHEN IT IS COOLED TO 273 C AND REDUCED TO A VOLUME OF 300 ML ? SOLUTION: (P1 x V1 ) / T1 = (P2 x V2 ) / T2, P1 = 5.0 ATM, V2 = 300 ML T1 = K, V1 = 850 ML T2 = ( ) = 546 K P2 = (P1 x V1 x T2 ) / (V2 x T1 ) P2 = (5.0 x x 546) / (300 x 1092) = 7.1 ATM

14
**SOLVING DALTON’S LAW PROBLEMS**

A TANK CONTAINS THREE GASES, N2 , Cl2 AND O2 . THE NITROGEN PRESSURE IS 2.0 ATM, THE CHLORINE 380 MM OF HG AND THE OXYGEN 5.0 ATM. WHAT IS THE PRESSURE IN THE TANK ? P TOTAL = PGAS A P GAS B P GAS C + P ……. P TOTAL = 2.0 ATM + (380/760)ATM ATM = 7.5 ATM OR x = 5700 MM OF HG

15
**SOLVING DALTON’S LAW PROBLEMS**

A TANK CONTAINS THREE GASES, H2 , Br2 AND O2 . THE MASS OF HYDROGEN IS 2.0 GRAMS, THE BROMINE MASS IS 240 GRAMS AND THE OXYGEN MASS IS 16.0 GRAMS. THE TOTAL PRESSURE IN THE TANK WAS 4.0 ATM. WHAT IS THE PRESSURE OF THE OXYGEN IN THE TANK ? SOLUTION: PGAS A = (N GAS A / N TOTAL) x PTOTAL MOLES = GRAMS / MOLAR MASS H2 = 2.0 / 2.0 = 1.0 MOLES, Br2 = 240 / 160 =1.5 MOLES O2 = 16.0 / 32 = 0.50 MOLES PO2 = ( (0.50) / ( )) x = ATM

16
**SOLVING IDEAL GAS LAW PROBLEMS**

WHAT IS THE TEMPERATURE OF 68.0 GRAMS OF OF HYDROGEN SULFIDE GAS WITH A VOLUME OF 6.0 LITERS AND A PRESSURE OF 5.0 ATMS ? SOLUTION: P x V = N x R x T, P = 5.0 ATM, V = 6.0 LITERS N = 68.0 / = 2.0 MOLES R = ATM x L / MOLES x K T = (P x V) / (N x R) T = (5.0 x 6.0) / (2.0 x ) = 183 K KELVIN = DEGREES CELSIUS = C0 , C0 = -90

17
**SOLVING IDEAL GAS LAW PROBLEMS**

WHAT IS THE VOLUME OF 64.0 GRAMS OF OF OXYGEN GAS WITH A TEMPERATURE OF 25 DEGREES CELSIUS AND A PRESSURE OF 3.0 ATMS ? SOLUTION: P x V = N x R x T, KELVIN = DEGREES T= = 298 K, P = 3.0 ATM N = 64.0 / = 2.0 MOLES R = ATM x L / MOLES x K V = (N x R x T) / P V = (2.0 x x 298) / 3.0 = C0 , C0 = -90

18
**SOLVING GRAHAM’S LAW PROBLEMS**

THE AVERAGE MOLECULAR SPEED OF AN OXYGEN MOLECULE AT A SPECIFIC TEMPERATURE IS 500 M/SEC. WHAT IS THE AVERAGE SPEED OF A NEON MOLECULE AT THE SAME TEMPERATURE ? v2 / v1 = ( m1 / m2)1/2 MOLAR MASS OXYGEN = 32.0, MOLAR MASS NEON = 20.0 m1 = 32, m2 = 20.0, v1 = 500 M / SEC v2 = ( m1 / m2)1/2 (v1) = (32.0/ 20.0) 1/2 (500) =632 M/SEC

19
**SOLVING GRAHAM’S LAW PROBLEMS**

WHAT IS MOLECULAR MASS OF SUBSTANCE X, IF AT A SPECIFIC TEMPERATURE THE SPEED OF A NITROGEN MOLECULE IS 800 M/SEC. THE AVERAGE SPEED OF THE UNKNOWN GAS IS 650 M/SEC AT THE SAME TEMPERATURE ? v2 / v1 = ( m1 / m2)1/2 (squaring both sides gives) m1 / m2 = v22 / v12 (solving for m1 gives) m1 = (v22 / v12) x m2 m2 = Molar mass of Nitrogen = 28.0 v2 = velocity of Nitrogen = 800 m / s v1 = velocity of unknown gas = 650 m /s m1 = (8002 / 6502) x 28 = 42.4

20
**SOLVING STP PROBLEMS Liters divide by 22.4 moles**

ONE MOLE OF ANY GAS AT STP ( 0 C OR 273 K AND 1 ATM OR 760 MM OF HG) OCCUPIES 22.4 LITERS HOW MANY GRAMS OF HYDROGEN ARE CONTAINED IN 89.6 LITERS AT STP ? SOLUTION : Liters divide by moles 89.6 / = 4.0 MOLES MOLES x MOLAR MASS = GRAMS 4.0 x = 8.0 GRAMS HYDROGEN

21
**SOLVING STP PROBLEMS 1.5 x 22.4 = 33.6 LITERS**

MOLE OF ANY GAS AT STP ( 0 C OR 273 K AND 1 ATM OR 760 MM OF HG) OCCUPIES 22.4 LITERS WHAT IS THE VOLUME OF 96.0 GRAMS OF SULFUR DIOXIDE AT STP ? SOLUTION : GRAMS / MOLAR MASS = MOLES 96.0 / = 1.5 MOLES MOLES X LITERs 1.5 x = LITERS

22
THE END

Similar presentations

Presentation is loading. Please wait....

OK

Gas Laws Practice Problems

Gas Laws Practice Problems

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on indian youth Ppt on polynomials in maths what does median Ppt on complex numbers class 11th result Ppt on the art of war Ppt on wild animals in india Ppt on conservation of momentum practice Ppt on tamper resistant prescription Download ppt on number system for class 10 Ppt on computer graphics 3d transformation Ppt on balanced diet